Vectors and Matrices I

A vector has both size and direction. A matrix is a rectangle of numbers. They look unrelated at first, and the connection only becomes clear later — for now, treat them as two sets of rules to get right.

Vectors and notation

A vector is drawn as an arrow: its length is the size, its direction is the way the arrow points. A quantity with size only — mass, temperature, time — is a scalar.

The same vector can be written three ways:

$$\overrightarrow{AB} \qquad \mathbf{a} \qquad \begin{pmatrix} 3 \\ 4 \end{pmatrix}$$

In the column vector form, the top number is the movement across and the bottom number is the movement up. So $\begin{pmatrix} 3 \\ 4 \end{pmatrix}$ means 3 right and 4 up, and $\begin{pmatrix} -2 \\ 5 \end{pmatrix}$ means 2 left and 5 up.

Column vector arithmetic

Add and subtract componentwise — tops with tops, bottoms with bottoms.

Let $\mathbf{a} = \begin{pmatrix} 2 \\ 3 \end{pmatrix}$ and $\mathbf{b} = \begin{pmatrix} 5 \\ -1 \end{pmatrix}$.

$$\mathbf{a} + \mathbf{b} = \begin{pmatrix} 2+5 \\ 3+(-1) \end{pmatrix} = \begin{pmatrix} 7 \\ 2 \end{pmatrix}$$

$$\mathbf{a} - \mathbf{b} = \begin{pmatrix} 2-5 \\ 3-(-1) \end{pmatrix} = \begin{pmatrix} -3 \\ 4 \end{pmatrix}$$

A scalar multiple multiplies both components:

$$3\mathbf{a} = \begin{pmatrix} 6 \\ 9 \end{pmatrix}$$

The result points the same way but is three times as long. A negative multiple reverses the direction.

So $\begin{pmatrix} 2 \\ 3 \end{pmatrix}$ and $\begin{pmatrix} 6 \\ 9 \end{pmatrix}$ are parallel, and so is $\begin{pmatrix} -4 \\ -6 \end{pmatrix}$ — it is $-2$ times the first, pointing the opposite way along the same line.

Magnitude

The magnitude (or modulus) of a vector is its length, written $|\mathbf{a}|$. It comes straight from Pythagoras:

$$\left|\begin{pmatrix} x \\ y \end{pmatrix}\right| = \sqrt{x^2 + y^2}$$

$$\left|\begin{pmatrix} 3 \\ 4 \end{pmatrix}\right| = \sqrt{9 + 16} = \sqrt{25} = 5$$

$$\left|\begin{pmatrix} -6 \\ 8 \end{pmatrix}\right| = \sqrt{36 + 64} = \sqrt{100} = 10$$

Position vectors

The position vector of a point is the vector from the origin to it. If $A$ is the point $(1,2)$ then $\overrightarrow{OA} = \begin{pmatrix} 1 \\ 2 \end{pmatrix}$.

To travel between two points, go back to the origin and out again:

$$\overrightarrow{AB} = \overrightarrow{OB} - \overrightarrow{OA}$$

$A$ is $(1,2)$ and $B$ is $(4,6)$. Find $\overrightarrow{AB}$ and its length.

$$\overrightarrow{AB} = \begin{pmatrix} 4 \\ 6 \end{pmatrix} - \begin{pmatrix} 1 \\ 2 \end{pmatrix} = \begin{pmatrix} 3 \\ 4 \end{pmatrix}$$

$$|\overrightarrow{AB}| = \sqrt{9+16} = 5$$

So $B$ is 3 right and 4 up from $A$, a straight-line distance of 5 units.

Matrices: order and addition

The order of a matrix is rows × columns, in that order.

$$\begin{pmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \end{pmatrix} \text{ has order } 2 \times 3$$

Two rows, three columns.

Addition and subtraction work element by element, and are only possible when the two matrices have the same order.

$$\begin{pmatrix} 2 & 1 \\ 3 & 4 \end{pmatrix} + \begin{pmatrix} 1 & 5 \\ 0 & 2 \end{pmatrix} = \begin{pmatrix} 3 & 6 \\ 3 & 6 \end{pmatrix}$$

Scalar multiplication multiplies every element:

$$2\begin{pmatrix} 2 & 1 \\ 3 & 4 \end{pmatrix} = \begin{pmatrix} 4 & 2 \\ 6 & 8 \end{pmatrix}$$

Matrix multiplication

Multiplication is the one that needs care, and it starts with a compatibility check.

$$(m \times n) \times (n \times p) = (m \times p)$$

The inner numbers must match, and the outer numbers give the order of the answer. So a $2 \times 3$ times a $3 \times 2$ works and gives a $2 \times 2$; a $2 \times 3$ times a $2 \times 3$ does not work at all.

To find each element, take a row from the left matrix and a column from the right, multiply the pairs and add.

$$A = \begin{pmatrix} 2 & 1 \\ 3 & 4 \end{pmatrix} \qquad B = \begin{pmatrix} 1 & 5 \\ 0 & 2 \end{pmatrix}$$

$$AB = \begin{pmatrix} (2)(1)+(1)(0) & (2)(5)+(1)(2) \\ (3)(1)+(4)(0) & (3)(5)+(4)(2) \end{pmatrix} = \begin{pmatrix} 2 & 12 \\ 3 & 23 \end{pmatrix}$$

Order matters here as it did with composite functions:

$$BA = \begin{pmatrix} (1)(2)+(5)(3) & (1)(1)+(5)(4) \\ (0)(2)+(2)(3) & (0)(1)+(2)(4) \end{pmatrix} = \begin{pmatrix} 17 & 21 \\ 6 & 8 \end{pmatrix}$$

$AB \ne BA$.